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Circular Motion and Gravitation · Solutions圆周运动与万有引力 · 详解

Companion to the Practice Set · Mark-by-mark walkthroughs · AP-Feeder / ON / BC / AB styles练习题配套答案 · 逐分讲解 · AP 衔接 / 安 / 卑 / 阿省考风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 🇨🇦 AB阿 AP-style MCQAP 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 AB Diploma-style阿尔伯塔毕业考风格 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解AP MCQ + ON/BC short answer · 18 marksAP 选择题 + 安/卑省考短答 · 共 18 分

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §1 Uniform circular motion匀速圆周运动 · HS-PS2-1 [3 marks][3 分]

Ball on a string, $r = 0.80$ m, $5.0$ rev/s. Speed of the ball?绳上的球，$r = 0.80$ m，每秒 $5.0$ 圈。球的速率？

Answer:答案：  (A)  $25\ \text{m/s}$

(a) Speed from circumference per revolution times frequency速率 = 每圈周长乘以频率 M1·A1·A1

In one revolution the ball travels the circumference $2\pi r$; it does $f = 5.0$ revolutions each second, so每转一圈球走过周长 $2\pi r$；每秒转 $f = 5.0$ 圈，故 $$ v \;=\; 2\pi r f \;=\; 2(3.14)(0.80)(5.0) \;=\; 25 \;\text{m/s}. $$ Option (A).选 (A)。
Why the distractors fail.干扰项分析。
(B) $4.0\ \text{m/s}$: uses $r \cdot f = 0.80 \times 5.0$ and forgets the factor $2\pi$ (omits the circumference).用 $r \cdot f = 0.80 \times 5.0$，漏掉了系数 $2\pi$（未用周长）。
(C) $0.16\ \text{m/s}$: computes $r/f = 0.80/5.0$, inverting the role of frequency.算成 $r/f = 0.80/5.0$，把频率的作用颠倒了。
(D) $13\ \text{m/s}$: uses $\pi r f$, dropping the factor of $2$.用 $\pi r f$，漏掉了因子 $2$。
Frequency, period and speed are three views of the same circular motion.频率、周期与速率是同一圆周运动的三种视角。 The three are linked by $T = 1/f$ and $v = 2\pi r / T = 2\pi r f$. The single most common error is leaving out the $2\pi$, because students confuse the radius with the distance travelled per revolution. The ball does not travel $r$ per revolution; it travels the full circumference $2\pi r$. A quick reasonableness check: a ball whirling five times per second on a $0.80$ m radius covers $2\pi(0.80) \approx 5.0$ m each lap, so $5.0\ \text{m} \times 5.0\ \text{s}^{-1} = 25$ m/s, confirming option (A).三者由 $T = 1/f$ 与 $v = 2\pi r / T = 2\pi r f$ 相联系。最常见的错误是漏掉 $2\pi$，因为学生把半径与每圈走过的距离混淆。球每圈走的不是 $r$，而是整个周长 $2\pi r$。快速合理性检验：半径 $0.80$ m、每秒五圈的球每圈走 $2\pi(0.80) \approx 5.0$ m，故 $5.0\ \text{m} \times 5.0\ \text{s}^{-1} = 25$ m/s，确认选项 (A)。
Q2EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §2 Centripetal force concept向心力概念 · HS-PS2-1 [3 marks][3 分]

Car on a flat circular curve at constant speed. Which real force provides the centripetal force?车在平坦圆弯道上匀速行驶。哪个真实力提供向心力？

Answer:答案：  (A)  static friction静摩擦力

(a) Identify the inward-pointing real force from a free-body diagram由受力图找出指向圆心的真实力 M1·A1·A1

On a flat curve the forces on the car are: gravity (down), the normal force (up, balancing gravity), and friction (horizontal). The net force must point toward the centre of the curve (horizontal, inward). Only friction has a horizontal component, so static friction supplies the centripetal force. Option (A).在平坦弯道上，车受的力有：重力（向下）、法向力（向上，平衡重力）、摩擦力（水平）。合力必须指向弯道圆心（水平、向内）。只有摩擦力有水平分量，故静摩擦力提供向心力。选 (A)。
Why the distractors fail.干扰项分析。
(B): "centripetal force" is never a new, separate force; it is the name for whatever real force happens to point inward."向心力"绝非新的独立力；它只是恰好指向圆心的某个真实力的名称。
(C): "centrifugal force" is a fictitious (apparent) force seen only in a rotating frame; it does not exist in an inertial frame and never appears on a correct FBD."离心力"是只在旋转参考系中出现的虚拟（表观）力；在惯性系中不存在，正确受力图上绝不出现。
(D): on a flat road the normal force is vertical, balancing gravity; it has no inward horizontal component.在平坦路面上法向力竖直，平衡重力；没有向内的水平分量。
"Centripetal" labels a role, not a new force."向心"标示的是一种角色，而非新的力。 The single biggest conceptual hurdle in circular motion is that "centripetal force" is not an extra force you add to a free-body diagram. It is whichever real force (tension, gravity, normal force, or friction) happens to point toward the centre. The correct procedure is always: draw the FBD with only real forces, find the resultant pointing inward, then set that resultant equal to $mv^2/r$. The fictitious outward "centrifugal force" should never appear in an inertial frame; invoking it is the classic mistake examiners look for.圆周运动中最大的概念障碍在于："向心力"不是你额外加到受力图上的力。它是恰好指向圆心的那个真实力（张力、重力、法向力或摩擦力）。正确步骤始终是：画出只含真实力的受力图，找出指向圆心的合力，再令该合力等于 $mv^2/r$。虚拟的向外"离心力"在惯性系中绝不应出现；用到它正是阅卷人寻找的经典错误。
Q3EASY易 🇨🇦 ON安 ON Provincial-style安大略省考风格 §2 Centripetal acceleration & force向心加速度与向心力 · SPH4U B2 [4 marks][4 分]

$1200$ kg car, flat curve $r = 80$ m, $v = 20$ m/s. (a) $a_c$. (b) Net force. (c) Direction and source.$1200$ kg 车，平坦弯道 $r = 80$ m，$v = 20$ m/s。(a) $a_c$。(b) 净力。(c) 方向与来源。

Answer:答案：  (a) $a_c = 5.0\ \text{m/s}^2$  ·  (b) $F_c = 6000\ \text{N}$  ·  (c) toward the centre; static friction指向圆心；静摩擦力

(a) Centripetal acceleration $a_c = v^2/r$向心加速度 $a_c = v^2/r$ M1·A1

$$ a_c \;=\; \frac{v^2}{r} \;=\; \frac{(20)^2}{80} \;=\; \frac{400}{80} \;=\; 5.0 \;\text{m/s}^2. $$

(b) Net force $F_c = ma_c$净力 $F_c = ma_c$ A1

$$ F_c \;=\; ma_c \;=\; (1200)(5.0) \;=\; 6000 \;\text{N.} $$

(c) Direction and source of the force力的方向与来源 A1

The net force points horizontally toward the centre of the curve. On a flat road it is supplied by the static friction between the tires and the road surface.净力沿水平方向指向弯道圆心。在平坦路面上，它由轮胎与路面之间的静摩擦力提供。
$a_c \propto v^2$: the inward acceleration grows with the square of the speed.$a_c \propto v^2$：向内加速度随速率的平方增长。 Because $a_c = v^2/r$, doubling the speed quadruples the centripetal acceleration and hence the friction the road must supply. This is why curves carry posted speed limits: at $40$ m/s instead of $20$ m/s the same car would need $24\,000$ N of friction, which the tires cannot deliver, and the car skids. Note also that $a_c$ depends only on $v$ and $r$ (not mass), but the force $F_c = ma_c$ does scale with mass: a heavier vehicle needs proportionally more friction.由于 $a_c = v^2/r$，速率加倍会使向心加速度变为四倍，从而所需摩擦力也变为四倍。这正是弯道设限速的原因：若车速为 $40$ m/s 而非 $20$ m/s，同一辆车需要 $24\,000$ N 摩擦力，轮胎无法提供，车便会打滑。还要注意：$a_c$ 只依赖 $v$ 与 $r$（与质量无关），但力 $F_c = ma_c$ 确实随质量增大：更重的车需要成比例更多的摩擦力。
Q4MEDIUM中 🇺🇸 US美 AP-style MCQAP 风格选择题 §4 Inverse-square law平方反比律 · HS-PS2-4 [3 marks][3 分]

Gravitational force is $F$. Distance tripled (masses unchanged). New force?引力为 $F$。距离变为三倍（质量不变）。新引力？

Answer:答案：  (A)  $F/9$

(a) Apply the inverse-square ratio套用平方反比比值 M1·A1·A1

Newton's law gives $F \propto 1/r^2$. Forming the ratio of new to old force with $r_{\text{new}} = 3r$:牛顿定律给出 $F \propto 1/r^2$。取新旧力之比，$r_{\text{新}} = 3r$： $$ \frac{F_{\text{new}}}{F} \;=\; \left(\frac{r}{r_{\text{new}}}\right)^2 \;=\; \left(\frac{r}{3r}\right)^2 \;=\; \frac{1}{9} \;\Longrightarrow\; F_{\text{new}} = \frac{F}{9}. $$ Option (A).选 (A)。
Why the distractors fail.干扰项分析。
(B) $F/3$: treats gravity as $\propto 1/r$ (linear) rather than $\propto 1/r^2$.把引力当作 $\propto 1/r$（线性），而非 $\propto 1/r^2$。
(C) $3F$: multiplies by $r$ instead of dividing by $r^2$ (gets the direction of the effect backward).乘以 $r$ 而非除以 $r^2$（效果方向弄反了）。
(D) $9F$: multiplies by $r^2$; correct factor magnitude but wrong direction.乘以 $r^2$；倍数大小正确但方向相反。
Inverse-square: a factor change in distance becomes its square in the force, inverted.平方反比：距离的倍数变化在力中变为其平方，且方向相反。 The defining feature of gravity (and of any field spreading over the surface of an expanding sphere) is $F \propto 1/r^2$. The safe technique is to write the ratio $F_2/F_1 = (r_1/r_2)^2$ explicitly rather than reasoning verbally, because the squaring trips up students who only scale linearly. Triple the distance and the force drops to one-ninth; halve the distance and it quadruples. The same $1/r^2$ law governs gravitational field strength $g$, light intensity, and electrostatic force, so this reasoning recurs across the whole physics curriculum.引力（以及任何在膨胀球面上扩散的场）的决定性特征是 $F \propto 1/r^2$。稳妥的方法是显式写出比值 $F_2/F_1 = (r_1/r_2)^2$，而非口头推理，因为平方会让只做线性缩放的学生出错。距离变三倍，力降为九分之一；距离减半，力变四倍。同样的 $1/r^2$ 律支配引力场强 $g$、光强和静电力，故这一推理贯穿整个物理课程。
Q5MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §3 Flat curve & friction平坦弯道与摩擦力 · Physics 12 [5 marks][5 分]

Flat curve $r = 50$ m, $\mu_s = 0.60$, $g = 9.8\ \text{m/s}^2$. (a) Show $v_{\max} = \sqrt{\mu_s g r}$. (b) Compute. (c) Mass dependence.平坦弯道 $r = 50$ m，$\mu_s = 0.60$，$g = 9.8\ \text{m/s}^2$。(a) 证明 $v_{\max} = \sqrt{\mu_s g r}$。(b) 计算。(c) 是否依赖质量。

Answer:答案：  (a) derivation推导  ·  (b) $v_{\max} \approx 17\ \text{m/s}$  ·  (c) independent of mass与质量无关

(a) Set maximum friction equal to the required centripetal force令最大摩擦力等于所需向心力 M1·A1

On a flat curve the normal force balances gravity, so $N = mg$. The maximum static friction is $f_{\max} = \mu_s N = \mu_s mg$, and this friction supplies the centripetal force:在平坦弯道上法向力平衡重力，故 $N = mg$。最大静摩擦力为 $f_{\max} = \mu_s N = \mu_s mg$，该摩擦力提供向心力： $$ \mu_s mg \;=\; \frac{mv_{\max}^2}{r} \;\Longrightarrow\; v_{\max}^2 \;=\; \mu_s g r \;\Longrightarrow\; v_{\max} \;=\; \sqrt{\mu_s g r}. $$

(b) Compute the maximum speed计算最大速率 M1·A1

$$ v_{\max} \;=\; \sqrt{(0.60)(9.8)(50)} \;=\; \sqrt{294} \;\approx\; 17 \;\text{m/s}. $$

(c) Mass dependence是否依赖质量 A1

The mass $m$ cancels from both sides in part (a), so the maximum speed does not depend on the mass of the car. A heavier car has more friction available, but it also needs proportionally more centripetal force, and the two effects cancel exactly.在 (a) 中质量 $m$ 从两边消去，故最大速率不依赖于车的质量。更重的车可获得更大摩擦力，但同时也需要成比例更多的向心力，两种效应恰好抵消。
Mass cancels whenever a force proportional to $m$ supplies the centripetal force.只要提供向心力的力正比于 $m$，质量就会相消。 Both the available friction ($\mu_s mg$) and the required centripetal force ($mv^2/r$) scale with mass, so $m$ disappears. The same cancellation makes the minimum loop speed $\sqrt{gr}$ and the orbital speed $\sqrt{GM/r}$ independent of the moving object's mass. The result is counter-intuitive but physically robust: the maximum cornering speed of a loaded truck and an empty one are the same on identical tires. The way to raise $v_{\max}$ is to bank the curve, increase $\mu_s$ (better tires), or increase $r$ (a gentler bend), never to change the vehicle's mass.可用摩擦力（$\mu_s mg$）与所需向心力（$mv^2/r$）都正比于质量，故 $m$ 消失。同样的相消使最低圈速 $\sqrt{gr}$ 和轨道速度 $\sqrt{GM/r}$ 都与运动物体的质量无关。这一结果反直觉但物理上稳健：在相同轮胎下，满载卡车和空载卡车的最大过弯速率相同。要提高 $v_{\max}$，应让弯道倾斜、增大 $\mu_s$（更好的轮胎）或增大 $r$（更平缓的弯），而绝非改变车的质量。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  简答题 · 详解AP-feeder FRQ + honors · 34 marksAP 衔接简答题 + 荣誉级 · 共 34 分

Section B · Worked SolutionsB 部分 · 详细解答

Q6MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §2 Centripetal force (ball on string)向心力（绳上的球） · HS-PS2-1 [8 marks][8 分]

$0.25$ kg ball, horizontal circle $r = 0.50$ m, $v = 4.0$ m/s. (a) $a_c$. (b) Tension. (c) Tension when $v$ doubled. (d) Motion if string breaks.$0.25$ kg 球，水平圆 $r = 0.50$ m，$v = 4.0$ m/s。(a) $a_c$。(b) 张力。(c) 速度加倍时的张力。(d) 绳断后的运动。

Answer:答案：  (a) $a_c = 32\ \text{m/s}^2$  ·  (b) $T = 8.0\ \text{N}$  ·  (c) $T = 32\ \text{N}$ ($\times 4$)  ·  (d) straight line, tangent to the circle沿圆的切线方向直线飞出

(a) Centripetal acceleration $a_c = v^2/r$向心加速度 $a_c = v^2/r$ M1·A1

$$ a_c \;=\; \frac{v^2}{r} \;=\; \frac{(4.0)^2}{0.50} \;=\; \frac{16}{0.50} \;=\; 32 \;\text{m/s}^2. $$

(b) Tension equals the centripetal force张力等于向心力 M1·A1

The string tension is the only horizontal force, so it provides the entire centripetal force:绳张力是唯一的水平力，故它提供全部向心力： $$ T \;=\; F_c \;=\; ma_c \;=\; (0.25)(32) \;=\; 8.0 \;\text{N.} $$

(c) New tension when the speed doubles速度加倍时的新张力 M1·A1

Since $T \propto v^2$ at fixed $r$, doubling the speed multiplies the tension by $2^2 = 4$:在 $r$ 不变时 $T \propto v^2$，速度加倍使张力变为 $2^2 = 4$ 倍： $$ T_{\text{new}} \;=\; \frac{m(2v)^2}{r} \;=\; 4 \times 8.0 \;=\; 32 \;\text{N.} $$

(d) Motion after the string breaks绳断后的运动 A1·A1

With the tension gone there is no net horizontal force, so by Newton's first law the ball travels in a straight line along the tangent to the circle at the point of release (at $4.0$ m/s, before gravity curves its path downward).张力消失后水平方向无合力，由牛顿第一定律，球沿释放点处圆的切线方向做直线运动（以 $4.0$ m/s，在重力使其路径向下弯曲之前）。
Identify the real inward force first, then set it equal to $mv^2/r$.先找出真实的向内力，再令其等于 $mv^2/r$。 The disciplined procedure is: (1) draw the free-body diagram, (2) name the real force pointing toward the centre (here the tension), (3) set it equal to $mv^2/r$. Part (c) showcases the quadratic dependence $F_c \propto v^2$: the single most-tested relationship in this topic. Doubling the speed quadruples the force, which is why strings snap and cars skid at "only slightly higher" speeds. Part (d) is the conceptual payoff: a centripetal force does not fling the ball outward; removing it lets the ball continue in a straight tangent line, exactly as Newton's first law predicts. There is no outward "centrifugal" force throwing it away from the centre.规范步骤是：(1) 画受力图，(2) 指出指向圆心的真实力（此处为张力），(3) 令其等于 $mv^2/r$。(c) 展示了二次依赖 $F_c \propto v^2$：本主题考查最多的关系。速度加倍使力变为四倍，这正是绳在"仅略高"的速度下断裂、车打滑的原因。(d) 是概念上的收获：向心力并不会把球甩向外侧；去掉它后球沿切线直线继续运动，恰如牛顿第一定律所预言。并不存在把球甩离圆心的向外"离心"力。
Q7MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §3 Vertical circles竖直圆 · SPH4U B [8 marks][8 分]

$500$ kg coaster car, vertical loop $r = 8.0$ m, $g = 9.8\ \text{m/s}^2$. (a) FBD + top equation. (b) Min top speed. (c) Bottom normal force at $14$ m/s. (d) As a multiple of weight.$500$ kg 过山车，竖直圆 $r = 8.0$ m，$g = 9.8\ \text{m/s}^2$。(a) 受力图与顶点方程。(b) 顶点最小速度。(c) $14$ m/s 时底部法向力。(d) 表示为车重的倍数。

Answer:答案：  (a) $mg = mv^2/r$  ·  (b) $v_{\min} \approx 8.9\ \text{m/s}$  ·  (c) $N \approx 17.2\ \text{kN}$  ·  (d) $N = 3.5\,mg$

(a) Free-body diagram and equation at the top顶点的受力图与方程 M1·A1

At the top of the loop both gravity ($mg$, down) and the normal force ($N$, down) point inward toward the centre. When the normal force is zero, gravity alone supplies the centripetal force:在圆圈顶部，重力（$mg$，向下）与法向力（$N$，向下）都指向圆心。当法向力为零时，重力单独提供向心力： $$ N + mg = \frac{mv^2}{r} \;\xrightarrow{\;N=0\;}\; mg = \frac{mv_{\min}^2}{r}. $$

(b) Minimum speed at the top顶点最小速度 M1·A1

$$ v_{\min} \;=\; \sqrt{gr} \;=\; \sqrt{(9.8)(8.0)} \;=\; \sqrt{78.4} \;\approx\; 8.9 \;\text{m/s.} $$

(c) Normal force at the bottom of the loop圆圈底部的法向力 M1·A1·A1

At the bottom, $N$ points up (inward) and $mg$ points down (outward), so $N - mg = mv^2/r$:在底部，$N$ 向上（向内），$mg$ 向下（向外），故 $N - mg = mv^2/r$： $$ N \;=\; mg + \frac{mv^2}{r} \;=\; (500)(9.8) + \frac{(500)(14)^2}{8.0} \;=\; 4900 + 12250 \;=\; 17\,150 \;\text{N} \approx 17.2 \;\text{kN.} $$

(d) Normal force as a multiple of weight法向力相当于车重的倍数 A1

The car's weight is $mg = 4900$ N, so $N / mg = 17\,150 / 4900 = 3.5$. The track pushes up with $3.5$ times the car's weight, and the riders feel $3.5$ times heavier than normal: the "apparent weight" increase at the bottom of a loop.车重为 $mg = 4900$ N，故 $N / mg = 17\,150 / 4900 = 3.5$。轨道向上的推力为车重的 $3.5$ 倍，乘客感觉比平时重 $3.5$ 倍：这就是圆圈底部的"表观重力"增大。
In a vertical circle, redo the free-body diagram at every position.在竖直圆中，每个位置都要重画受力图。 The defining feature of a vertical circle is that the relationship between gravity and the normal force changes with position. At the top both point inward, so they add: $N + mg = mv^2/r$. At the bottom they oppose, so they subtract: $N - mg = mv^2/r$. Using the same equation at both points is the single most common error. The minimum-speed condition $v_{\min} = \sqrt{gr}$ is independent of mass (gravity provides exactly the needed centripetal force), but the bottom normal force scales with mass. Notice the answer $N = 3.5\,mg$ is what produces the "stomach-drop" sensation: it is not extra gravity, it is the seat pushing up harder to curve your path.竖直圆的决定性特征是重力与法向力的关系随位置变化。在顶部两者都向内，故相加：$N + mg = mv^2/r$。在底部两者相反，故相减：$N - mg = mv^2/r$。在两点使用同一个方程是最常见的错误。最小速度条件 $v_{\min} = \sqrt{gr}$ 与质量无关（重力恰好提供所需向心力），但底部法向力随质量增大。注意答案 $N = 3.5\,mg$ 正是"心往下沉"感觉的来源：那不是额外的重力，而是座椅更用力地向上推以弯曲你的运动路径。
Q8HARD难 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §4-§5 Universal gravitation & fields万有引力与引力场 · Physics 12 [8 marks][8 分]

Planet $M = 6.4 \times 10^{23}$ kg, $R = 3.4 \times 10^6$ m, $G = 6.674 \times 10^{-11}$. (a) Surface $g$. (b) Weight of 50 kg. (c) $g$ at altitude $h = R$.行星 $M = 6.4 \times 10^{23}$ kg，$R = 3.4 \times 10^6$ m，$G = 6.674 \times 10^{-11}$。(a) 表面 $g$。(b) 50 kg 的重力。(c) 高度 $h = R$ 处的 $g$。

Answer:答案：  (a) $g \approx 3.7\ \text{N/kg}$  ·  (b) $W \approx 185\ \text{N}$  ·  (c) $g_h \approx 0.92\ \text{N/kg}$

(a) Surface gravitational field $g = GM/R^2$表面引力场 $g = GM/R^2$ M1·A1·A1

$$ g \;=\; \frac{GM}{R^2} \;=\; \frac{(6.674\times10^{-11})(6.4\times10^{23})}{(3.4\times10^6)^2} \;=\; \frac{4.27\times10^{13}}{1.156\times10^{13}} \;\approx\; 3.7 \;\text{N/kg.} $$

(b) Weight $W = mg$重力 $W = mg$ M1·A1

$$ W \;=\; mg \;=\; (50)(3.69) \;\approx\; 185 \;\text{N.} $$

(c) Field at altitude $h = R$ using the inverse-square ratio用平方反比比值求高度 $h = R$ 处的场强 M1·A1·A1

At altitude $h = R$ the centre-to-centre distance is $r = R + R = 2R$. Since $g \propto 1/r^2$:在高度 $h = R$ 处，质心距为 $r = R + R = 2R$。因 $g \propto 1/r^2$： $$ g_h \;=\; g\left(\frac{R}{2R}\right)^2 \;=\; \frac{g}{4} \;=\; \frac{3.69}{4} \;\approx\; 0.92 \;\text{N/kg.} $$
Altitude is not the distance $r$; the inverse-square ratio beats recomputing.高度不是距离 $r$；用平方反比比值胜过从头重算。 Two lessons live in this problem. First, the gravitational field strength $g = GM/R^2$ carries units of N/kg, numerically equal to m/s² (the surface acceleration). Second, and the single most-tested trap in gravitation: $r$ is the centre-to-centre distance, never the altitude. "At an altitude equal to one radius" means $r = 2R$, not $r = R$. Once the surface value is known, the inverse-square ratio $g_h = g\,(R/r)^2$ gives the altitude value in one line, avoiding a fresh full calculation and the chance of an exponent slip. Doubling $r$ quarters the field, which is why a $3.7$ N/kg surface field falls to $0.92$ N/kg.本题包含两点教训。第一，引力场强 $g = GM/R^2$ 单位为 N/kg，数值上等于 m/s²（表面加速度）。第二，也是引力中考查最多的陷阱：$r$ 是质心到质心的距离，绝非高度。"在等于一个半径的高度处"意味着 $r = 2R$，而非 $r = R$。一旦已知表面值，平方反比比值 $g_h = g\,(R/r)^2$ 一行即可给出高空值，避免重新完整计算和指数出错的风险。$r$ 加倍使场强变为四分之一，故 $3.7$ N/kg 的表面场降至 $0.92$ N/kg。
Q9HARD难Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §7 Satellites & orbits卫星与轨道 · HS-PS2-4 (above two-body floor)（超出两体基准） [10 marks][10 分]

ISS at altitude $h = 400$ km. $G = 6.674\times10^{-11}$, $M_E = 5.97\times10^{24}$ kg, $R_E = 6.37\times10^6$ m. (a) State $r$. (b) Show $v = \sqrt{GM_E/r}$. (c) Orbital speed. (d) Period (min). (e) Heavier satellite, same radius.国际空间站，高度 $h = 400$ km。$G = 6.674\times10^{-11}$，$M_E = 5.97\times10^{24}$ kg，$R_E = 6.37\times10^6$ m。(a) 写出 $r$。(b) 证明 $v = \sqrt{GM_E/r}$。(c) 轨道速度。(d) 周期（分钟）。(e) 同半径上质量更大的卫星。

Answer:答案：  (a) $r = 6.77\times10^6\ \text{m}$  ·  (b) derivation推导  ·  (c) $v \approx 7670\ \text{m/s}$  ·  (d) $T \approx 92\ \text{min}$  ·  (e) same speed速度相同

(a) Orbital radius (centre-to-centre)轨道半径（质心到质心） A1

$$ r \;=\; R_E + h \;=\; 6.37\times10^6 + 4.00\times10^5 \;=\; 6.77\times10^6 \;\text{m.} $$ This is the orbital radius, not the altitude $400$ km.这是轨道半径，而非高度 $400$ km。

(b) Orbital speed from gravity $=$ centripetal force由引力 $=$ 向心力求轨道速度 M1·A1

Gravity supplies the centripetal force for the orbiting satellite of mass $m$:引力为质量 $m$ 的绕行卫星提供向心力： $$ \frac{GmM_E}{r^2} \;=\; \frac{mv^2}{r} \;\Longrightarrow\; v^2 \;=\; \frac{GM_E}{r} \;\Longrightarrow\; v \;=\; \sqrt{\frac{GM_E}{r}}. $$

(c) Calculate the orbital speed计算轨道速度 M1·A1·A1

$$ v \;=\; \sqrt{\frac{(6.674\times10^{-11})(5.97\times10^{24})}{6.77\times10^6}} \;=\; \sqrt{5.89\times10^7} \;\approx\; 7670 \;\text{m/s.} $$

(d) Orbital period轨道周期 M1·A1

$$ T \;=\; \frac{2\pi r}{v} \;=\; \frac{2\pi(6.77\times10^6)}{7670} \;\approx\; 5545 \;\text{s} \;\approx\; 92 \;\text{min.} $$

(e) A four-times-heavier satellite at the same radius同一半径上质量为四倍的卫星 A1·A1

Its speed is the same. In part (b) the satellite mass $m$ cancelled, so $v = \sqrt{GM_E/r}$ depends only on $r$ and the central mass $M_E$. Quadrupling the satellite mass does not change its orbital speed (or period).它的速度相同。在 (b) 中卫星质量 $m$ 相消，故 $v = \sqrt{GM_E/r}$ 只依赖 $r$ 与中心质量 $M_E$。卫星质量变为四倍不改变其轨道速度（或周期）。
For an orbit, gravity is the centripetal force, and the satellite's own mass cancels out.对于轨道，引力就是向心力，卫星自身质量相消。 Setting $GmM_E/r^2 = mv^2/r$ is the master move for every orbit problem: the satellite mass $m$ appears on both sides and cancels, leaving $v = \sqrt{GM_E/r}$ and $T = 2\pi\sqrt{r^3/(GM_E)}$. This is why a tiny CubeSat and the $420$-tonne ISS share the same orbital speed at $400$ km. A reasonableness check: orbital speeds are kilometres per second ($\approx 7.7$ km/s here), and low-Earth-orbit periods are roughly $90$ minutes, so the ISS circles Earth about 16 times a day. The recurring trap is using the altitude for $r$ instead of $R_E + h$; always write $r$ explicitly first, as in part (a).令 $GmM_E/r^2 = mv^2/r$ 是每个轨道问题的核心步骤：卫星质量 $m$ 在两边出现并相消，留下 $v = \sqrt{GM_E/r}$ 与 $T = 2\pi\sqrt{r^3/(GM_E)}$。这正是为何微小的立方星与 $420$ 吨的国际空间站在 $400$ km 处轨道速度相同。合理性检验：轨道速度量级为千米每秒（此处约 $7.7$ km/s），近地轨道周期约 $90$ 分钟，故国际空间站每天绕地约 $16$ 圈。反复出现的陷阱是把高度当作 $r$ 而非 $R_E + h$；务必像 (a) 那样先显式写出 $r$。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模与应用 · 详解AB Diploma + Universal · 28 marks阿省毕业考 + 通用题型 · 共 28 分

Section C · Modeling and ApplicationsC 部分 · 建模与应用

Q10MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §6 Kepler's Third Law开普勒第三定律 · 20-C1.7k [9 marks][9 分]

Planet Kestrel orbits a Sun-like star at $4\times$ Earth's orbital radius. Earth's period is $1.0$ yr. (a) State Kepler's Third Law as a ratio. (b) Kestrel's period. (c) Radius of a planet with period $27$ yr.行星 Kestrel 绕类太阳恒星运行，半径为地球的 $4$ 倍。地球周期 $1.0$ 年。(a) 写出开普勒第三定律比值形式。(b) Kestrel 的周期。(c) 周期 $27$ 年的行星半径。

Answer:答案：  (a) $T_1^2/r_1^3 = T_2^2/r_2^3$  ·  (b) $T = 8.0\ \text{yr}$  ·  (c) $r = 9.0\,r_E$

(a) Kepler's Third Law as a ratio开普勒第三定律的比值形式 M1·A1·A1

For any two bodies orbiting the same central mass, $T^2/r^3$ is the same constant $4\pi^2/(GM)$. Equating the two:对绕同一中心质量运行的任意两个天体，$T^2/r^3$ 为同一常数 $4\pi^2/(GM)$。令两者相等： $$ \frac{T_1^2}{r_1^3} \;=\; \frac{T_2^2}{r_2^3}. $$ Because the constant $4\pi^2/(GM)$ is identical for both bodies, it cancels when we take the ratio, so $G$ and the star's mass $M$ never need to be known: only the ratio of radii matters.由于常数 $4\pi^2/(GM)$ 对两个天体相同，取比值时被消去，故无需知道 $G$ 与恒星质量 $M$：只有半径之比重要。

(b) Period of Kestrel ($r_K = 4r_E$)Kestrel 的周期（$r_K = 4r_E$） M1·A1·A1

$$ \left(\frac{T_K}{T_E}\right)^2 \;=\; \left(\frac{r_K}{r_E}\right)^3 \;=\; 4^3 \;=\; 64 \;\Longrightarrow\; \frac{T_K}{T_E} \;=\; \sqrt{64} \;=\; 8. $$ $$ T_K \;=\; 8 \times 1.0 \;=\; 8.0 \;\text{yr.} $$

(c) Radius of a planet with $T = 27$ yr周期 $T = 27$ 年的行星半径 M1·A1·A1

$$ \left(\frac{r}{r_E}\right)^3 \;=\; \left(\frac{T}{T_E}\right)^2 \;=\; 27^2 \;=\; 729 \;\Longrightarrow\; \frac{r}{r_E} \;=\; 729^{1/3} \;=\; 9 \;\Longrightarrow\; r \;=\; 9.0\,r_E. $$
Kepler's Third Law as a ratio sidesteps $G$ and the central mass entirely.开普勒第三定律的比值形式完全绕开 $G$ 与中心质量。 When two objects orbit the same star, writing $T_1^2/r_1^3 = T_2^2/r_2^3$ is far faster than computing each period from $T = 2\pi\sqrt{r^3/(GM)}$, because the messy constant $4\pi^2/(GM)$ cancels. The relationship is $T^2 \propto r^3$, so a factor change in radius becomes that factor to the power $3/2$ in the period (and the inverse for the reverse direction). Four times the radius gives $4^{3/2} = 8$ times the period; nine times the radius would give $27$ times the period, which is exactly the inverse of part (c). This is the reasoning Newton used to confirm that the same inverse-square gravity governing falling apples also holds the planets in their orbits.当两个天体绕同一恒星运行时，写出 $T_1^2/r_1^3 = T_2^2/r_2^3$ 远比用 $T = 2\pi\sqrt{r^3/(GM)}$ 分别计算周期更快，因为繁琐的常数 $4\pi^2/(GM)$ 被消去。关系式为 $T^2 \propto r^3$，故半径的倍数变化在周期中变为该倍数的 $3/2$ 次幂（反方向则取倒数）。半径四倍给出 $4^{3/2} = 8$ 倍周期；半径九倍则给出 $27$ 倍周期，这恰好是 (c) 的逆运算。这正是牛顿用来确认支配落地苹果的同一平方反比引力也维系行星轨道的推理。
Q11MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §4-§5 Surface gravity (applied)表面重力（应用） · 20-B2.5k [9 marks][9 分]

Exoplanet $M = 8.0\times10^{24}$ kg, $R = 7.0\times10^6$ m, $G = 6.674\times10^{-11}$. (a) Derive $g = GM/R^2$. (b) Surface $g$. (c) Weight of 60 kg lander. (d) Heavier or lighter than on Earth?系外行星 $M = 8.0\times10^{24}$ kg，$R = 7.0\times10^6$ m，$G = 6.674\times10^{-11}$。(a) 推导 $g = GM/R^2$。(b) 表面 $g$。(c) 60 kg 着陆器的重力。(d) 比在地球更重还是更轻？

Answer:答案：  (a) $g = GM/R^2$  ·  (b) $g \approx 10.9\ \text{m/s}^2$  ·  (c) $W \approx 654\ \text{N}$  ·  (d) heavier更重

(a) Derive surface gravity from $mg = GmM/R^2$由 $mg = GmM/R^2$ 推导表面重力 M1·A1

At the surface, the weight $mg$ equals the gravitational force $GmM/R^2$. The test mass $m$ cancels:在地表，重力 $mg$ 等于引力 $GmM/R^2$。试探质量 $m$ 相消： $$ mg \;=\; \frac{GmM}{R^2} \;\Longrightarrow\; g \;=\; \frac{GM}{R^2}. $$

(b) Calculate the surface gravity计算表面重力 M1·A1·A1

$$ g \;=\; \frac{(6.674\times10^{-11})(8.0\times10^{24})}{(7.0\times10^6)^2} \;=\; \frac{5.34\times10^{14}}{4.9\times10^{13}} \;\approx\; 10.9 \;\text{m/s}^2. $$

(c) Weight of the lander着陆器的重力 M1·A1

$$ W \;=\; mg \;=\; (60)(10.9) \;\approx\; 654 \;\text{N.} $$

(d) Compared with Earth与地球相比 A1·A1

The lander weighs more here. The surface gravity $10.9\ \text{m/s}^2$ exceeds Earth's $9.8\ \text{m/s}^2$, so the same $60$ kg mass weighs more than its $\approx 588$ N on Earth.着陆器在此处更重。表面重力 $10.9\ \text{m/s}^2$ 大于地球的 $9.8\ \text{m/s}^2$，故同样 $60$ kg 的质量比在地球上的 $\approx 588$ N 更重。
Surface gravity depends on both mass and radius: $g = GM/R^2$, not on $M$ alone.表面重力同时取决于质量与半径：$g = GM/R^2$，而非仅取决于 $M$。 A planet can be more massive than Earth yet have weaker surface gravity if it is also much larger, because $g$ falls with $R^2$. Here the exoplanet has more mass and a similar radius to Earth, so its surface gravity is higher. The derivation in part (a) shows why $g$ is "the same for all objects": the falling mass $m$ cancels, so a feather and a hammer accelerate identically (Apollo 15 demonstrated this on the airless Moon). Always carry the powers of ten carefully: a single exponent slip between $10^{-11}$ and $10^{24}$ changes the answer by orders of magnitude. A reasonableness check: surface gravities of rocky worlds run a few m/s², so $10.9$ m/s² is plausible for an Earth-mass, Earth-sized planet.一颗行星可以比地球质量更大却表面重力更弱，只要它也大得多，因为 $g$ 随 $R^2$ 减小。此处系外行星质量更大、半径与地球相近，故表面重力更高。(a) 的推导说明了为何 $g$ "对所有物体相同"：下落质量 $m$ 相消，故羽毛与锤子加速度相同（阿波罗 15 号在无空气的月球上演示了这一点）。务必谨慎处理十次幂：在 $10^{-11}$ 与 $10^{24}$ 之间差一个指数，答案就会差几个数量级。合理性检验：岩质天体的表面重力为几 m/s²，故对地球质量、地球大小的行星而言 $10.9$ m/s² 是合理的。
Q12HARD难Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §7 Mass from orbital data由轨道数据求质量 · HS-PS2-4 (above two-body floor)（超出两体基准） [10 marks][10 分]

Moon orbits a planet, $r = 1.9\times10^7$ m, $T = 1.5$ days, $G = 6.674\times10^{-11}$. (a) Derive $M = 4\pi^2 r^3/(GT^2)$. (b) Period in seconds. (c) Planet mass. (d) Why moon mass absent.卫星绕行星运行，$r = 1.9\times10^7$ m，$T = 1.5$ 天，$G = 6.674\times10^{-11}$。(a) 推导 $M = 4\pi^2 r^3/(GT^2)$。(b) 周期（秒）。(c) 行星质量。(d) 为何卫星质量缺席。

Answer:答案：  (a) $M = 4\pi^2 r^3/(GT^2)$  ·  (b) $T = 1.296\times10^5\ \text{s}$  ·  (c) $M \approx 2.4\times10^{23}\ \text{kg}$  ·  (d) moon mass cancels卫星质量相消

(a) Derive the planet's mass from gravity $=$ centripetal force由引力 $=$ 向心力推导行星质量 M1·A1·A1

Let the moon have mass $m$. Gravity supplies the centripetal force, and the orbital speed is $v = 2\pi r / T$:设卫星质量为 $m$。引力提供向心力，轨道速度为 $v = 2\pi r / T$： $$ \frac{GmM}{r^2} \;=\; \frac{mv^2}{r} \;=\; \frac{m}{r}\left(\frac{2\pi r}{T}\right)^2 \;\Longrightarrow\; \frac{GM}{r^2} \;=\; \frac{4\pi^2 r}{T^2} \;\Longrightarrow\; M \;=\; \frac{4\pi^2 r^3}{G T^2}. $$

(b) Convert the period to seconds将周期换算为秒 A1

$$ T \;=\; 1.5 \times 86\,400 \;=\; 1.296\times10^5 \;\text{s.} $$

(c) Calculate the planet's mass计算行星质量 M1·A1·A1·A1

$$ M \;=\; \frac{4\pi^2 (1.9\times10^7)^3}{(6.674\times10^{-11})(1.296\times10^5)^2} \;=\; \frac{2.71\times10^{23}}{(6.674\times10^{-11})(1.68\times10^{10})}. $$ $$ M \;=\; \frac{2.71\times10^{23}}{1.12} \;\approx\; 2.4\times10^{23} \;\text{kg.} $$

(d) Why the moon's mass does not appear为何卫星质量不出现 A1·A1

In part (a) the moon's mass $m$ appears on both sides of $GmM/r^2 = mv^2/r$ and cancels. The orbital motion is set entirely by the central mass $M$, the radius $r$, and the period $T$; the orbiting body's own mass plays no role. This is exactly why measuring a moon's $r$ and $T$ lets astronomers weigh the planet it orbits.在 (a) 中卫星质量 $m$ 出现在 $GmM/r^2 = mv^2/r$ 两边并相消。轨道运动完全由中心质量 $M$、半径 $r$ 和周期 $T$ 决定；绕行天体自身的质量不起作用。这正是为何测量卫星的 $r$ 与 $T$ 就能让天文学家"称量"它所绕的行星。
Kepler's Third Law rearranged is a cosmic scale: orbital $r$ and $T$ weigh the central body.变形的开普勒第三定律是一台宇宙秤：轨道 $r$ 与 $T$ 可称量中心天体。 Rearranging $T^2 = 4\pi^2 r^3/(GM)$ to solve for $M$ is how the mass of the Sun, of Jupiter, and of any planet with an observable moon is measured. The orbiting body's mass cancels, so a satellite's $r$ and $T$ alone determine the central mass. Watch two pitfalls: (1) cube the radius and square the period before dividing, keeping the powers of ten organised; (2) convert the period to SI seconds first ($1.5$ days $= 1.296\times10^5$ s) since $G$ is in SI units. A reasonableness check: the result $\approx 2.4\times10^{23}$ kg is a few percent of Earth's mass ($5.97\times10^{24}$ kg), plausible for a small planet or a large moon.将 $T^2 = 4\pi^2 r^3/(GM)$ 变形求解 $M$，正是测量太阳质量、木星质量乃至任何拥有可观测卫星的行星质量的方法。绕行天体的质量相消，故仅凭卫星的 $r$ 与 $T$ 即可确定中心质量。注意两个陷阱：(1) 相除前先把半径立方、周期平方，并理清十次幂；(2) 先把周期换算为国际单位制的秒（$1.5$ 天 $= 1.296\times10^5$ s），因为 $G$ 用国际单位制。合理性检验：结果 $\approx 2.4\times10^{23}$ kg 约为地球质量（$5.97\times10^{24}$ kg）的百分之几，对小行星或大卫星而言是合理的。